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A particle moves in a straight line, so ...

A particle moves in a straight line, so that after `t` second, the distance `x` from a fixed point `O` on the line is given by `x=(l-2)^(2)(t-5)`. Then

A

after `2s` velocity of particle is zero

B

after `2s`, the particle reaches at `O`

C

the acceleation is negative when `t lt 3s`

D

All of the above

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Knowledge Check

  • If a particle moving in a straight line and its distance x cms from a fixed point O on the line is given by x=sqrt(1+t^(2)) cms, then acceleration of the particle at t sec. is

    A
    `(1)/(x^(2))" cm/sec"^(2)`
    B
    `(-1)/(x^(2))" cm/sec"^(2)`
    C
    `(1)/(x^(3))" cm/sec"^(2)`
    D
    `(-1)/(x^(3))" cm/sec"^(2)`

  • A particle is moving on a straight line and its distance x cms from a fixed point O on the line is given by x=sqrt(t^(2)+1) then the velocity of particle at t=1 is

    A
    `1/(sqrt(2))`
    B
    `1/(sqrt(3))`
    C
    `1/(2sqrt(2))`
    D
    `1/(3sqrt(2))`

  • Speed v of a particle moving along a straight line, when it is a distance X from a fixed point on the line is given by V^(2)=108-9X^(2) (all quantities in S.I. unit). Then

    A
    The motion is uniformly accelereted along the straight line
    B
    The magnitude of the accelerated at a distance `3 cm` from the fixed point is `0.27 m//s^(2)`.
    C
    The motion is harminic about `x=sqrt(12) m`.
    D
    The maximum displacement from the fixed point is `4 cm`.

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